Problem: Simplify and expand the following expression: $ \dfrac{5r + 3}{5r + 10}-\dfrac{r - 6}{4r + 5} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(5r + 10)(4r + 5)$ Multiply the first term by $\dfrac{4r + 5}{4r + 5}$ $ \begin{align*} \dfrac{5r + 3}{5r + 10} \times \dfrac{4r + 5}{4r + 5} & = \dfrac{(5r + 3)(4r + 5)}{(5r + 10)(4r + 5)} \\ & = \dfrac{20r^2 + 37r + 15}{(5r + 10)(4r + 5)}\end{align*} $ Multiply the second term by $\dfrac{5r + 10}{5r + 10}$ $ \begin{align*} \dfrac{r - 6}{4r + 5} \times \dfrac{5r + 10}{5r + 10} & = \dfrac{(r - 6)(5r + 10)}{(4r + 5)(5r + 10)} \\ & = \dfrac{5r^2 - 20r - 60}{(4r + 5)(5r + 10)}\end{align*} $ Now we have: $ = \dfrac{20r^2 + 37r + 15}{(5r + 10)(4r + 5)} - \dfrac{5r^2 - 20r - 60}{(4r + 5)(5r + 10)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{20r^2 + 37r + 15 - (5r^2 - 20r - 60)}{(5r + 10)(4r + 5)} $ $ = \dfrac{20r^2 + 37r + 15 - 5r^2 + 20r + 60}{(5r + 10)(4r + 5)} $ $ = \dfrac{15r^2 + 57r + 75}{(5r + 10)(4r + 5)}$ Expand the denominator: $ = \dfrac{15r^2 + 57r + 75}{20r^2 + 65r + 50}$